Kyle is flying a helicopter and it is rising at 6.0 m/s when he releases a bag. After 2.0 s

(a) what is the bag's velocity?
-13.6 m/s
(b) how far has the bag fallen?

(c) how far below the helicopter is the bag?

To calculate the bag's velocity, we need to use the concept of freefall acceleration. The acceleration due to gravity is approximately 9.8 m/s^2. Since the bag is dropping, its velocity will be negative.

(a) To find the bag's velocity, we can use the formula:

velocity = initial velocity + (acceleration × time)

Given:
initial velocity (u) = 0 m/s (since the bag was released)
acceleration (a) = acceleration due to gravity = -9.8 m/s^2 (negative because the bag is dropping)
time (t) = 2.0 s

velocity = 0 + (-9.8 × 2.0) = -19.6 m/s

Therefore, the bag's velocity is -19.6 m/s.

(b) To find how far the bag has fallen, we can use the equation of motion:

distance fallen = (initial velocity × time) + (0.5 × acceleration × time^2)

Given:
initial velocity (u) = 0 m/s (since the bag was released)
acceleration (a) = acceleration due to gravity = -9.8 m/s^2 (negative because the bag is dropping)
time (t) = 2.0 s

distance fallen = (0 × 2.0) + (0.5 × -9.8 × 2.0^2)
= 0 + (0.5 × -9.8 × 4.0)
= 0 + (-19.6) = -19.6 m

Therefore, the bag has fallen 19.6 meters.

(c) To find how far below the helicopter the bag is, we need to consider the initial height from where the bag was released. If no information is given about the initial height, we cannot determine this value.

To solve this problem, we can use the equations of motion. The key concept to understand here is that when the bag is released, it will continue to move vertically downwards due to the force of gravity acting on it.

(a) To find the bag's velocity after 2.0 seconds, we need to consider its initial velocity, the acceleration due to gravity, and the time it takes to fall.

Since the helicopter is rising at 6.0 m/s, the initial velocity of the bag is also 6.0 m/s in the upward direction (since the bag is released while moving upwards with the helicopter). However, once the bag is released, it will start to fall downwards.

The acceleration due to gravity is approximately 9.8 m/s² and acts in the downward direction.

Using the equation of motion:
Vf = Vi + at

where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and t is the time, we can calculate the bag's final velocity.

Vf = 6.0 m/s + (-9.8 m/s²) × 2.0 s
Vf = 6.0 m/s - 19.6 m/s
Vf = -13.6 m/s

Therefore, the bag's velocity after 2.0 seconds is -13.6 m/s (downwards).

(b) To find how far the bag has fallen, we can use the formula for distance covered under constant acceleration:

d = Vi × t + 0.5 × a × t²

where d is the distance, Vi is the initial velocity, a is the acceleration, and t is the time.

Since the bag is released with an initial velocity of 6.0 m/s and it falls downward with an acceleration of 9.8 m/s², we can substitute these values into the equation:

d = 6.0 m/s × 2.0 s + 0.5 × (-9.8 m/s²) × (2.0 s)²
d = 12.0 m - 19.6 m
d = -7.6 m

Therefore, the bag has fallen 7.6 meters in the 2.0 seconds.

(c) To find how far below the helicopter the bag is, we need to consider the displacement in the downward direction.

The displacement is always given by:

displacement = Vi × t + 0.5 × a × t²

Since the bag is falling downwards, the displacement will be negative. So, using the same values for Vi, a, and t:

displacement = 6.0 m/s × 2.0 s + 0.5 × (-9.8 m/s²) × (2.0 s)²
displacement = 12.0 m - 19.6 m
displacement = -7.6 m

Therefore, the bag is located 7.6 meters below the helicopter.